【原创】leetCodeOj --- Intersection of Two Linked Lists 解题报告(经典的相交链表找交点)

题目地址：
https://oj.leetcode.com/problems/intersection-of-two-linked-lists/
题目内容：

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
↘
c1 → c2 → c3
↗
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

If the two linked lists have no intersection at all, return

null

.

The linked lists must retain their original structure after the function returns.

You may assume there are no cycles anywhere in the entire linked structure.

Your code should preferably run in O(n) time and use only O(1) memory.

方法：

首先，既然是时间复杂度O(n)，那么就不能一个一个点那样试；

其次，既然空间复杂度要求O(1)，既然就不能用stack或者unordered_map之类的数据结构来找链表交点。

那么，我们需要一个比较酷炫的trick来找交点。

先转化问题：

假设有A、B两条链表相交，请求出交点到A链表头结点的距离。（所谓距离，就是头结点走几次能到）

先看具体求法：

0、计算A链表的长度lenA

1、计算B链表的长度lenB

2、逆转A链表（关键）

3、重新计算B链表的长度newLenB

4、返回result = (newLenB - lenB + lenA - 1) / 2

具体到这道题，还需要把A链表又逆转回来，因为你不能更改链表原来的结构。然后重新读取链表，返回第result个结点就OK了。

那么这具体求法究竟是怎么来的？

自己动手试试就明白了。其实就是算出A链表结点在交点旁边的分布数，画图太麻烦了，如果有时间再补一个，或者谁不理解回复一下我就补

全部代码：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if (headA == NULL || headB == NULL)
return NULL;
int addressA; // fin of A.
int addressB; // fin of B.
int lenA = countLength(headA,&addressA);
int lenB = countLength(headB,&addressB);
if (addressA != addressB) // if has a intersect
return NULL;
ListNode *tmpHeadA = (ListNode *)addressA; // to store headA's tail for reverse.
reverseLink(headA);
int newLenB = countLength(headB,&addressB);
int toNew = findCount(lenA,lenB,newLenB);
reverseLink(tmpHeadA);
return findNthNode(headA,toNew);
}

int findCount(int lenA,int lenB,int newLenB)
{
int gap = newLenB - lenB;
return (gap + lenA - 1) / 2;
}

ListNode *findNthNode(ListNode *head,int toN)
{
int count = 0;
while (toN != count)
{
head = head->next;
count ++;
}
return head;
}

int countLength(ListNode *head,int *fin)
{
int count = 0;
while (head)
{
*fin = (int)head;
head = head->next;
count ++;
}
return count;
}

void reverseLink(ListNode *head)
{
ListNode *pre = NULL;
ListNode *now = head;
ListNode *nxt = head->next;
while (1)
{
now->next = pre;
pre = now;
now = nxt;
if (nxt)
nxt = nxt->next;
else
break;
}
}
};